Case III - Unknown Mass - The Balance Principle
Since we know that the meter stick was balanced, the positive torques (ccw) exactly cancel out the negative torques (cw) of the system.
Since the torque is given by the mN, the following formula describes the balance of the torques:
(m1)(g)(x1) = (m2)(g)(x2)
Dividing by 9.81m/s2 from both sides yields the following equation which will be used for simplicity's sake:
(m1)(x1) = (m2)(x2) where m1 = unknown mass, m2 = 199.95g = 0.2kg, x1 = distance from fulcrum of m1, x2 = distance from fulcrum of m2
Mass D
(m1)(0.33 = (0.2)(0.5)
m1 = 0.303kg
Mass F
(m1)(0.33) = (0.2)(0.39)
m1 =0.236kg
Mass R
(m1)(0.33) = (0.2)(0.294)
m1 = 0.178kg
Mass B
(m1)(0.33) = (0.2)(0.471)
m1 = 0.290kg
Mass E
(m1)(0.33) = (0.2)(0.403)
m1 = 0.244kg
The approximate mass of the hanger that kept the m1 suspended on the meter stick was 0.024kg,
After subtracting 0.024kg from each m1 yields the following masses for each hanging mass (accepted masses for each also listed):
D = 0.279kg 0.27102kg
F = 0.212kg 0.19670kg
R = 0.154kg 0.13240kg
B = 0.266kg 0.25751kg
E = 0.220kg 0.20647kg
Percent errors for each mass are given below as given by the following formula:
(Actual - expected) / expected
D = 2.952%
F = 7.614%
R = 16.667%
B = 3.101%
E = 6.796%
Case IV - Meter Stick with One Mass
Since the torque is given by the mN, the following formula describes the balance of the torques:
(m1)(g)(x1) = (m2)(g)(x2)
Dividing by 9.81m/s2 from both sides yields the following equation which will be used for simplicity's sake:
(m1)(x1) = (m2)(x2)
Substituting 99.44g = 0.099kg in for m1, 0.263m in for x1 (since x1 is placed at 0cm on the meter stick and the fulcrum was measured at 0.237m when balanced), and 0.237m in for x2 (since 0.5m - 0.263m = 0.237m and all of the mass of the meter stick can be considered a point mass at the center of the meter stick) yields:
(0.1)(0.263) = (0.237)(m2)
Solving for m2 yields 0.111kg
Actual mass of the meter stick was measured to be 0.110kg
Running these numbers through the percent error formula yields a percent error of 0.909%.
Discussion Questions
1) Small forces can produce little to no torque while small forces can produce larger torques because torque is the product of both force and the length of the moment arm. Since T(tau)=rF, a small force can product a large torque by being applied through a lengthy lever arm. Similarly, a large force may produce a small torque if the moment arm is extremely short.
2) The support force of the fulcrum is not used in the calculations of the torque of the system because the amount of force exerted upward on the meter stick by the fulcrum is exactly cancelled out by the downward force of the meter stick on the fulcrum. This essentially means they are force pairs that add no additional force to the system since there was no acceleration so by definition of Newton's second law, no acceleration equates to zero force. In addition, since the fulcrum is balanced exactly where the center of mass is, any force that could be applied would be translational motion not rotational.
3) Since the torque is given by the mN, the following formula describes the balance of the torques:
(m1)(g)(x1) = (m2)(g)(x2) + (m3)(g)(x3)
Dividing by 9.81m/s2 from both sides yields the following equation which will be used for simplicity's sake:
(m1)(x1) = (m2)(x2) + (m3)(x3)
Substituting the appropriate values into the equation as given by the lab yields:
(0.22)(0.35) = (0.1)(Mms) + (0.5)(0.12)
Solving for Mms (mass of the meter stick) yields a mass of 0.17kg.
Case III - Unknown Mass - The Balance Principle
Since we know that the meter stick was balanced, the positive torques (ccw) exactly cancel out the negative torques (cw) of the system.
Since the torque is given by the mN, the following formula describes the balance of the torques:
(m1)(g)(x1) = (m2)(g)(x2)
Dividing by 9.81m/s2 from both sides yields the following equation which will be used for simplicity's sake:
(m1)(x1) = (m2)(x2) where m1 = unknown mass, m2 = 199.95g = 0.2kg, x1 = distance from fulcrum of m1, x2 = distance from fulcrum of m2
Mass D
(m1)(0.33 = (0.2)(0.5)
m1 = 0.303kg
Mass F
(m1)(0.33) = (0.2)(0.39)
m1 =0.236kg
Mass R
(m1)(0.33) = (0.2)(0.294)
m1 = 0.178kg
Mass B
(m1)(0.33) = (0.2)(0.471)
m1 = 0.290kg
Mass E
(m1)(0.33) = (0.2)(0.403)
m1 = 0.244kg
The approximate mass of the hanger that kept the m1 suspended on the meter stick was 0.024kg,
After subtracting 0.024kg from each m1 yields the following masses for each hanging mass (accepted masses for each also listed):
D = 0.279kg 0.27102kg
F = 0.212kg 0.19670kg
R = 0.154kg 0.13240kg
B = 0.266kg 0.25751kg
E = 0.220kg 0.20647kg
Percent errors for each mass are given below as given by the following formula:
(Actual - expected) / expected
D = 2.952%
F = 7.614%
R = 16.667%
B = 3.101%
E = 6.796%
Case IV - Meter Stick with One Mass
Since the torque is given by the mN, the following formula describes the balance of the torques:
(m1)(g)(x1) = (m2)(g)(x2)
Dividing by 9.81m/s2 from both sides yields the following equation which will be used for simplicity's sake:
(m1)(x1) = (m2)(x2)
Substituting 99.44g = 0.099kg in for m1, 0.263m in for x1 (since x1 is placed at 0cm on the meter stick and the fulcrum was measured at 0.237m when balanced), and 0.237m in for x2 (since 0.5m - 0.263m = 0.237m and all of the mass of the meter stick can be considered a point mass at the center of the meter stick) yields:
(0.1)(0.263) = (0.237)(m2)
Solving for m2 yields 0.111kg
Actual mass of the meter stick was measured to be 0.110kg
Running these numbers through the percent error formula yields a percent error of 0.909%.
Discussion Questions
1) Small forces can produce little to no torque while small forces can produce larger torques because torque is the product of both force and the length of the moment arm. Since T(tau)=rF, a small force can product a large torque by being applied through a lengthy lever arm. Similarly, a large force may produce a small torque if the moment arm is extremely short.
2) The support force of the fulcrum is not used in the calculations of the torque of the system because the amount of force exerted upward on the meter stick by the fulcrum is exactly cancelled out by the downward force of the meter stick on the fulcrum. This essentially means they are force pairs that add no additional force to the system since there was no acceleration so by definition of Newton's second law, no acceleration equates to zero force. In addition, since the fulcrum is balanced exactly where the center of mass is, any force that could be applied would be translational motion not rotational.
3) Since the torque is given by the mN, the following formula describes the balance of the torques:
(m1)(g)(x1) = (m2)(g)(x2) + (m3)(g)(x3)
Dividing by 9.81m/s2 from both sides yields the following equation which will be used for simplicity's sake:
(m1)(x1) = (m2)(x2) + (m3)(x3)
Substituting the appropriate values into the equation as given by the lab yields:
(0.22)(0.35) = (0.1)(Mms) + (0.5)(0.12)
Solving for Mms (mass of the meter stick) yields a mass of 0.17kg.